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Let f:[0,1]→ R be such that f(xy)=f(x).f(y), for all x,y∈ [0,1], and f(0)≠ 0. If y=y(x) satisfies the differential equation, dy dx=f(x) with y(0)=1…

JEE Main 2019 — Mathematics Algebra

2019mcqeasy

Let f:[0,1]→Rf:[0,1]\rightarrow Rf:[0,1]→R be such that f(xy)=f(x).f(y),f(xy)=f(x).f(y),f(xy)=f(x).f(y), for all x,y∈[0,1],x,y\in [0,1],x,y∈[0,1], and f(0)≠0.f(0)\neq 0.f(0)=0. If y=y(x)y=y(x)y=y(x) satisfies the differential equation, dydx=f(x)\frac{dy}{dx}=f(x)dxdy​=f(x) with y(0)=1y(0)=1y(0)=1 then y(14)+y(34)y(\frac{1}{4})+y(\frac{3}{4})y(41​)+y(43​) is equal to:

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    555

  2. B

    222

  3. C

    333

  4. D

    444

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